$A$ and $B$ are two independent events such that $P\left( A \cup B^{'} \right) = 0.8$ and $P(A) = 0.3.$ Then, $P(B)$ is

Question

Dr. Abubacker · Accepted Answer

(a)
$P\left( A \cup B^{'} ight) = P(A) + P\left( B^{'} ight) - P(A)P(B^{'})$
$	herefore 0.8 = 0.3 + P\left( B^{'} ight) - 0.3P(B')$
$\Rightarrow 0.5 = P\left( B^{'} ight)(0.7)$
$\Rightarrow P\left( B^{'} ight) = \frac{5}{7}$
$	herefore P(B) = 1 - \frac{5}{7} = \frac{2}{7}$

Q. $A$ and $B$ are two independent events such that $P\left( A \cup B^{'} \right) = 0.8$ and $P(A) = 0.3.$ Then, $P(B)$ is

Explanation

Reviewed by Dr. Abubacker

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